Question

A paperboard measuring 29 in long and 21 in wide has a semicede cut out of it, as shown below. What is the perimeter of the paperboard after the semicirde is removed? (Use the value 3.14 for pi, and do not round your answer. Be sure to include the unit your answer)

Answer 1

To get the perimeter of the figure shown, we have to add up the lenght of its sides.

Notice that we have 3 linear segments, measuring 29", 21" and 29". Adding up this segments, we get:

`29in+21in+29in=79in`

Now, we have one segment remaining: a semicircle.

The radius of such semicircle is half of the lenght of the 21" sides. This is 10.5"

The perimeter of the semicircle is one half of the perimeter of a full circumference:

`(2\pi r)/(2)\rightarrow\pi r`

Therefore, the perimeter of this semicircle is:

`\pi r\rightarrow(3.14)\cdot(10.5in)\rightarrow32.97in`

Adding this up with the other 79" we've already calculated, we'll find the perimeter of the whole figure:

`79in+32.97in=111.97in`

Therefore, we can conclude that: